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Question

Find the number of terms of the AP 121,117,113,,3 ?

A
32
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B
30
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C
28
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D
26
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Solution

The correct option is A 32

Given

121,117,113...,3

a=121

d=117121=4

an=3

an=a+(n1)d

3=121+(n1)(4)

3=1214n+4

4n=128

n=1284

n=32



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